Problem: You have found the following ages (in years) of all $4$ zebras at your local zoo: $ 7,\enspace 1,\enspace 9,\enspace 14$ What is the average age of the zebras at your zoo? What is the variance? Round your answers to the nearest tenth. Average age: $ $
Explanation: Because we have data for all $4$ zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$. To find the population mean, add up the values of all $4$ ages and divide by $4$. $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 1 + 9 + 14}{{4}} = {7.75\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-0.75$ years $0.56$ years $^2$ $1$ year $-6.75$ years $45.56$ years $^2$ $9$ years $1.25$ years $1.56$ years $^2$ $14$ years $6.25$ years $39.06$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean, we can find the population variance $({\sigma^2})$, without introducing any bias, by simply averaging the squared deviations from the mean: $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.56} + {45.56} + {1.56} + {39.06}} {{4}} $ $ {\sigma^2} = \dfrac{{86.74}}{{4}} = {21.69\text{ years}^2} $ The average zebra at the zoo is $7.8$ years old. The population variance is $21.7$ years $^2$.